Question

A researcher compares the effectiveness of two different instructional methods for teaching physiology. A sample of 226 students using Method 1 produces a testing average of 69.8. A sample of 191 students using Method 2 produces a testing average of 81.5. Assume that the population standard deviation for Method 1 is 11, while the population standard deviation for Method 2 is 16.55. Determine the 99% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2. Step 1 of 3: Find the point estimate for the true difference between the population means. A researcher compares the effectiveness of two different instructional methods for teaching physiology. A sample of 226 students using Method 1 produces a testing average of 69.8. A sample of 191 students using Method 2 produces a testing average of 81.5. Assume that the population standard deviation for Method 1 is 11, while the population standard deviation for Method 2 is 16.55. Determine the 99% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

Step 2 of 3: Calculate the margin of error of a confidence interval for the difference between the two population means. Round your answer to six decimal places. A researcher compares the effectiveness of two different instructional methods for teaching physiology. A sample of 226 students using Method 1 produces a testing average of 69.8. A sample of 191 students using Method 2 produces a testing average of 81.5. Assume that the population standard deviation for Method 1 is 11, while the population standard deviation for Method 2 is 16.55. Determine the 99% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

Step 3 of 3: Construct the 99% confidence interval. Round your answers to one decimal place.

Answer 1

1)

Point estimate = 69.8 - 81.5 = -11.7

2)

sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(121/226 + 273.9025/191)
sp = 1.4034

Given CI level is 0.99, hence α = 1 - 0.99 = 0.01                  
α/2 = 0.01/2 = 0.005,zc = z(α/2, df) = 2.576                  
                  
Margin of Error                  
ME = zc * sp                  
ME = 2.58 * 1.4034                  
ME = 3.620772                  

3)

  
CI = (x1bar - x2bar - zc * sp , x1bar - x2bar + zc * sp)                  
CI = (69.8 - 81.5 - 2.576 * 1.4034 , 69.8 - 81.5 - 2.576 * 1.4034                  
CI = (-15.3 , -8.1)